3.81 \(\int \frac{(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=512 \[ \frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^4 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^4 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{5 e^3 \sqrt{e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{4 b^3 d (a+b \cos (c+d x))}-\frac{15 a e^4 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{4 b^4 d \sqrt{e \sin (c+d x)}}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \]

[Out]

(5*(3*a^2 - 2*b^2)*e^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*b^(7/2)*(-a
^2 + b^2)^(3/4)*d) + (5*(3*a^2 - 2*b^2)*e^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqr
t[e])])/(8*b^(7/2)*(-a^2 + b^2)^(3/4)*d) - (15*a*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(4*b
^4*d*Sqrt[e*Sin[c + d*x]]) + (5*a*(3*a^2 - 2*b^2)*e^4*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x
)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b^4*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (5*a*(3*a^2 - 2*
b^2)*e^4*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b^4*(a^2 - b*(
b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (5*e^3*(3*a + 2*b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(4*b^3*
d*(a + b*Cos[c + d*x])) + (e*(e*Sin[c + d*x])^(5/2))/(2*b*d*(a + b*Cos[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 1.13776, antiderivative size = 512, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {2693, 2863, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^4 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^4 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{5 e^3 \sqrt{e \sin (c+d x)} (3 a+2 b \cos (c+d x))}{4 b^3 d (a+b \cos (c+d x))}-\frac{15 a e^4 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{4 b^4 d \sqrt{e \sin (c+d x)}}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sin[c + d*x])^(7/2)/(a + b*Cos[c + d*x])^3,x]

[Out]

(5*(3*a^2 - 2*b^2)*e^(7/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*b^(7/2)*(-a
^2 + b^2)^(3/4)*d) + (5*(3*a^2 - 2*b^2)*e^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqr
t[e])])/(8*b^(7/2)*(-a^2 + b^2)^(3/4)*d) - (15*a*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(4*b
^4*d*Sqrt[e*Sin[c + d*x]]) + (5*a*(3*a^2 - 2*b^2)*e^4*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x
)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b^4*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (5*a*(3*a^2 - 2*
b^2)*e^4*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b^4*(a^2 - b*(
b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (5*e^3*(3*a + 2*b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(4*b^3*
d*(a + b*Cos[c + d*x])) + (e*(e*Sin[c + d*x])^(5/2))/(2*b*d*(a + b*Cos[c + d*x])^2)

Rule 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2863

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2867

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]))/((a_) + (b_.)*sin[(e_.) + (
f_.)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p, x], x] + Dist[(b*c - a*d)/b, Int[(g*Cos[e + f*x])^
p/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2702

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, -Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[(b*g)/f, Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
+ f*x]]*(q - b*Cos[e + f*x])), x], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^3} \, dx &=\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (5 e^2\right ) \int \frac{\cos (c+d x) (e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx}{4 b}\\ &=\frac{5 e^3 (3 a+2 b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 e^4\right ) \int \frac{-b-\frac{3}{2} a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{4 b^3}\\ &=\frac{5 e^3 (3 a+2 b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (15 a e^4\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{8 b^4}+\frac{\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac{1}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{8 b^4}\\ &=\frac{5 e^3 (3 a+2 b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2}}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2}}-\frac{\left (5 \left (3 a^2-2 b^2\right ) e^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{8 b^3 d}-\frac{\left (15 a e^4 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{8 b^4 \sqrt{e \sin (c+d x)}}\\ &=-\frac{15 a e^4 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 b^4 d \sqrt{e \sin (c+d x)}}+\frac{5 e^3 (3 a+2 b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (5 \left (3 a^2-2 b^2\right ) e^5\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{4 b^3 d}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2} \sqrt{e \sin (c+d x)}}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2} \sqrt{e \sin (c+d x)}}\\ &=-\frac{15 a e^4 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 b^4 d \sqrt{e \sin (c+d x)}}+\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}-\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b^4 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{5 e^3 (3 a+2 b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{8 b^3 \sqrt{-a^2+b^2} d}+\frac{\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{8 b^3 \sqrt{-a^2+b^2} d}\\ &=\frac{5 \left (3 a^2-2 b^2\right ) e^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}+\frac{5 \left (3 a^2-2 b^2\right ) e^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac{15 a e^4 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 b^4 d \sqrt{e \sin (c+d x)}}+\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b^4 \left (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}-\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b^4 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{5 e^3 (3 a+2 b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{4 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{5/2}}{2 b d (a+b \cos (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 16.6899, size = 1954, normalized size = 3.82 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Sin[c + d*x])^(7/2)/(a + b*Cos[c + d*x])^3,x]

[Out]

(((-a^2 + b^2)/(2*b^3*(a + b*Cos[c + d*x])^2) + (9*a)/(4*b^3*(a + b*Cos[c + d*x])))*Csc[c + d*x]^3*(e*Sin[c +
d*x])^(7/2))/d - ((e*Sin[c + d*x])^(7/2)*((14*a*Cos[c + d*x]^2*(a + b*Sqrt[1 - Sin[c + d*x]^2])*((a*(-2*ArcTan
[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]
])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c +
d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]))/(4*Sqr
t[2]*Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*
x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4,
Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (b^2
*Sin[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(
-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-1 + Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2))
+ (12*b*Cos[c + d*x]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*S
qrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)]
+ Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] - Log[Sqrt
[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(-a^2 + b^2)^(3/4)
+ (5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c +
d*x]])/(Sqrt[1 - Sin[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2
)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (-a
^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2
+ b^2*(-1 + Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2]) - (4*b*Cos[c + d*x]*Cos[2*(c +
d*x)]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c +
d*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 + ((1 + I)*Sqr
t[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + ((1/4 - I/4)*(-2*a^2 + b^2)*Log[S
qrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]])/(b^(3/2)*(-a^2 +
b^2)^(3/4)) - ((1/4 - I/4)*(-2*a^2 + b^2)*Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c
+ d*x]] + I*b*Sin[c + d*x]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + (4*Sqrt[Sin[c + d*x]])/b - (4*a*AppellF1[5/4, 1/2
, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(5/2))/(5*(a^2 - b^2)) + (10*a*(a^2
- b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]])/(Sqrt
[1 - Sin[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^
2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*Ap
pellF1[5/4, 3/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-1 +
Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - 2*Sin[c + d*x]^2)*Sqrt[1 - Sin[c + d*x]^2])))/(8*b^3*d*Sin[c +
d*x]^(7/2))

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Maple [B]  time = 14.263, size = 5404, normalized size = 10.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x)

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))**(7/2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(7/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out